I just don't understand why [itex]\displaystyle \frac{\sin x}{x} = c[/itex]
means that [itex] y = c [/itex] has to be tangent to
[itex]\displaystyle y = \frac{\sin x}{x}[/itex] at the second hump?

I'm not seeing the connection...

Thank you.

Look at the second graph. If y = c x is not tangent to the graph of y = sin(x) "near the second hump", then the graph of y = c x will intersect the graph of y = sin(x) in either 7 places, or in 3 places.

One way to solve this equation is graphically, i.e., where does the graph of y = c intercept the graph of [itex]\displaystyle y=\frac{\sin(x)}{x}\,?[/itex] I think the question to be asking is, "Should these two graphs be tangent at the point of intersection?"

If y = c is tangent to [itex]\displaystyle y=\frac{\sin(x)}{x}\,[/itex] near [itex]\displaystyle x=\frac{5\pi}{2}\,,[/itex] then it must be true that on either side of the point of tangency, [itex]\displaystyle c>\frac{\sin(x)}{x}\,.[/itex] It can be shown that this inequality holds.

Therefore, the slope of [itex]\displaystyle y=\frac{\sin(x)}{x}\,[/itex] is zero at the same value of x at which the the graph of y = c x is tangent to the graph of [itex]\displaystyle y=\sin(x)\,[/itex] near the second hump. Use this value of x to solve for c.